Find the area of parallelogram where one of its diagonals is of length 6.8 cm and perpendicular distance of this diagonal from the opposite vertex is of length 7.5 cm

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Solution

$Area of △ ABD = \frac{1}{2} \times BD \times AE = \frac{1}{2} \times 6.8 \times 7.5 {cm}^{2} = 25.5 {cm}^{2}$

We know that the diagonal of a parallelogram divides it into two triangles of equal area.
Area of $△$ BDC = area of $△$ ABD = 25.5 cm²
∴ Area of parallelogram ABCD = area of $△$ BDC + area of $△$ ABD
= (25.5 + 25.5) cm²
= 51 cm²

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Find the area of parallelogram where one of its diagonals is of length 6.8 cm and perpendicular distance of this diagonal from the opposite vertex is of length 7.5 cm

Area of △ABD=12×BD×AE = 12×6.8×7.5 cm2 = 25.5 cm2 We know that the diagonal of a parallelogram divides it into two triangles of equal area. Area of △BDC = area of △ABD = 25.5 cm2 ∴ Area of parallelogram ABCD = area of △BDC + area of △ABD = (25.5 + 25.5) cm2 = 51 cm2